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complement-of-base-10-integer | C++| 100% faster| EASY TO UNDERSTAND | clean and efficient | c-100-faster-easy-to-understand-clean-an-2chc | Please upvote to motivate me in my quest of documenting all leetcode solutions. HAPPY CODING:)\nAny suggestions and improvements are always welcome\n```\nclass | aarindey | NORMAL | 2021-06-14T16:12:09.733173+00:00 | 2021-06-14T16:12:09.733213+00:00 | 863 | false | ***Please upvote to motivate me in my quest of documenting all leetcode solutions. HAPPY CODING:)\nAny suggestions and improvements are always welcome***\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n int sum=0,i=0;\n if(n==0)\n return 1; \n while(n>0)\n {\n ... | 13 | 1 | ['C'] | 0 |
complement-of-base-10-integer | ✅Java - One Line Code, Easy Explanation!!! | java-one-line-code-easy-explanation-by-p-osuh | Idea\n\nBInary Complement of 5(101) is 2(010)\n\nSee,\nConvert all bits of 5(101) to ones i.e 111(5)\nNow we know the answer i.e 7(111)-5(101) =2(010) \n\nSo, f | pgthebigshot | NORMAL | 2022-01-04T05:39:57.020944+00:00 | 2022-01-04T06:29:50.739857+00:00 | 590 | false | **Idea**\n\nBInary Complement of 5(101) is 2(010)\n\nSee,\nConvert all bits of 5(101) to ones i.e 111(5)\nNow we know the answer i.e 7(111)-5(101) =2(010) \n\nSo, first find the size of bits of n i.e 3 in this case and pow(2,3)= 8 \nTherefore the answer be like 8-1-5=2\n\n````\nclass Solution {\n public int bitwiseC... | 9 | 4 | ['Java'] | 3 |
complement-of-base-10-integer | ✅ [Solution] Swift: Complement of Base 10 Integer (+ test cases) | solution-swift-complement-of-base-10-int-a29l | swift\nclass Solution {\n func bitwiseComplement(_ n: Int) -> Int {\n var s: Int = 1\n while s < n { s = (s << 1) | 1 }\n return s - n\n | AsahiOcean | NORMAL | 2021-05-27T19:47:22.671946+00:00 | 2022-01-04T03:38:02.915771+00:00 | 1,187 | false | ```swift\nclass Solution {\n func bitwiseComplement(_ n: Int) -> Int {\n var s: Int = 1\n while s < n { s = (s << 1) | 1 }\n return s - n\n }\n}\n```\n\n---\n\n<p>\n<details>\n<summary>\n<img src="https://git.io/JDblm" height="24">\n<b>TEST CASES</b>\n</summary>\n\n<p><pre>\n<b>Result:</b> Ex... | 9 | 0 | ['Swift'] | 0 |
complement-of-base-10-integer | Idiots solution for Python 3 (100%) no loops/bit manipulation | idiots-solution-for-python-3-100-no-loop-dwxd | Take the next power of 2 that is bigger than N, then subtract 1 from it to get the number that has the same leftmost bit, but with all of the bits to the right | fatmoron | NORMAL | 2020-05-04T10:23:12.100848+00:00 | 2020-05-04T10:23:12.100898+00:00 | 1,117 | false | Take the next power of 2 that is bigger than N, then subtract 1 from it to get the number that has the same leftmost bit, but with all of the bits to the right filled. \n\nSubtract N from this number and you will get the bits to the right that are 0 converted to 1, which is what you are looking for, e.g. 010 as input, ... | 9 | 1 | [] | 1 |
complement-of-base-10-integer | Java, simple 1 line bit manipulation (left & right shifts), 0ms | java-simple-1-line-bit-manipulation-left-c920 | Just actually do the complement operation ~, after first shifting away all the leading zeroes, and then shift away all the trailing ones.\n\n public int bitw | jpv | NORMAL | 2019-03-17T11:52:37.636362+00:00 | 2019-03-17T11:52:37.636405+00:00 | 944 | false | Just actually do the complement operation ~, after first shifting away all the leading zeroes, and then shift away all the trailing ones.\n```\n public int bitwiseComplement(int N) {\n return (N==0) ? 1 : (~(N<<(N=Integer.numberOfLeadingZeros(N))))>>N;\n }\n``` | 9 | 1 | [] | 3 |
complement-of-base-10-integer | Easy Python solution | easy-python-solution-by-vistrit-z5xa | \ndef bitwiseComplement(self, n: int) -> int:\n x=""\n for i in bin(n)[2:]:\n if i=="1":\n x+="0"\n else:\n | vistrit | NORMAL | 2022-01-04T01:20:47.226730+00:00 | 2022-01-04T01:20:47.226777+00:00 | 1,099 | false | ```\ndef bitwiseComplement(self, n: int) -> int:\n x=""\n for i in bin(n)[2:]:\n if i=="1":\n x+="0"\n else:\n x+="1"\n return int(x,2)\n``` | 8 | 1 | ['Python', 'Python3'] | 1 |
complement-of-base-10-integer | JAVA One Liner 100% Faster Solution || Explaination | java-one-liner-100-faster-solution-expla-8vcf | \n\nn = 10 decimal \nn = ...001010\n~n = ...110101\nIntege | abhishekpatel_ | NORMAL | 2021-07-06T04:14:03.448704+00:00 | 2021-07-06T04:31:36.993608+00:00 | 419 | false | \n```\nn = 10 decimal \nn = ...001010\n~n = ...110101\nInteger.highestOneBit(n) = ...001000\nInteger.highestOneBit(n)-1 = ...000111\n~n & Integer.highestOneBit(n)-1 = ...000101\n\n```\n\n\n\n\n*... | 8 | 0 | ['Java'] | 1 |
complement-of-base-10-integer | [Java/Python] Straightforward Bit Manipulation - Clean code - O(logN) | javapython-straightforward-bit-manipulat-3l6w | Complexity\n- Time: O(logN)\n- Space: O(1)\n\nPython\npython\nclass Solution(object):\n def bitwiseComplement(self, N):\n def setBit(x, k):\n | hiepit | NORMAL | 2020-10-05T07:38:59.097292+00:00 | 2020-10-05T07:58:41.582500+00:00 | 485 | false | **Complexity**\n- Time: `O(logN)`\n- Space: `O(1)`\n\n**Python**\n```python\nclass Solution(object):\n def bitwiseComplement(self, N):\n def setBit(x, k):\n return (1 << k) | x\n \n if N == 0: return 1\n i, ans = 0, 0\n while N > 0:\n if N % 2 == 0:\n ... | 8 | 6 | [] | 1 |
complement-of-base-10-integer | [Python] XOR masking explained | python-xor-masking-explained-by-sidheshw-4lal | For easy writing num is written as n\n\nSo the agenda of the question is to flip all the bits of given number. i.e., \nn = 4 (100) => 3 (011)\n\nOne way of ache | sidheshwar_s | NORMAL | 2022-01-04T04:14:45.557875+00:00 | 2022-01-04T04:14:45.557923+00:00 | 680 | false | *For easy writing num is written as n*\n\nSo the agenda of the question is to flip all the bits of given number. i.e., \n` n = 4 (100) => 3 (011)`\n\nOne way of acheving it is **xor between the number and its mask**, i.e., `100 ^ 111 == 011`\n\n* **So now how to get its mask ?**\n\t* A simple way would be to initializ... | 7 | 1 | ['Bit Manipulation', 'Python'] | 2 |
complement-of-base-10-integer | C++, 1-liner, 9 CPU instructions, and future (C++20) solution | c-1-liner-9-cpu-instructions-and-future-d7b1p | This is a complement of the other "standard solutions". If you haven\'t yet, go read the solution or other posts first, please.\n\nSince this solution uses eith | alvin-777 | NORMAL | 2020-10-06T05:39:41.404460+00:00 | 2020-10-06T05:39:41.404514+00:00 | 648 | false | > This is a complement of the other "standard solutions". If you haven\'t yet, go read the solution or other posts first, please.\n\nSince this solution uses either a C++20 function `std::countl_zero()` (at this point of time, it seems even the latest GCC (10.2) hasn\'t implement it, Microsoft added it in VS2019 versio... | 7 | 0 | ['C'] | 2 |
complement-of-base-10-integer | Best Solution | best-solution-by-kumar21ayush03-rvf4 | Approach\nUsing Bitwise Operator\n\n# Complexity\n- Time complexity:\nO(logn)\n\n- Space complexity:\nO(1)\n\n# Code\n\nclass Solution {\npublic:\n int bitwi | kumar21ayush03 | NORMAL | 2023-05-19T06:04:38.343732+00:00 | 2023-05-19T06:04:38.343774+00:00 | 689 | false | # Approach\nUsing Bitwise Operator\n\n# Complexity\n- Time complexity:\n$$O(logn)$$\n\n- Space complexity:\n$$O(1)$$\n\n# Code\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n int num = n;\n int mask = 0;\n if (n == 0)\n return 1;\n while (num != 0) {\n ... | 6 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | C++ XOR, simple solution | c-xor-simple-solution-by-apoorvtyagi10-onjg | Approach: \n\n- Find the first number which is greater than "num" and have all 1s (Eg: 1, 3(11), 7(111), 15(1111), ...)\n- The relation b/w numbers with all 1s | apoorvtyagi10 | NORMAL | 2021-12-27T13:16:47.371181+00:00 | 2021-12-29T19:20:21.640112+00:00 | 351 | false | Approach: \n\n- Find the first number which is greater than "num" and have all 1s (Eg: 1, 3(11), 7(111), 15(1111), ...)\n- The relation b/w numbers with all 1s is 2x + 1, starting from x=1\n- Finally xor the num with x.\n- If num = 5(binary 101), x evaluates as 7 (binary 111) and (num^x) evaluates as 010 which is compl... | 6 | 0 | ['C', 'C++'] | 1 |
complement-of-base-10-integer | Java O(1) | 1 line Solution | java-o1-1-line-solution-by-tbekpro-a4qn | I find the length of binary representation. For example, for 5 it will base-2 is "101" and length of this String is 3. \n2. Then by doing 1 << 3, I will have "1 | tbekpro | NORMAL | 2022-12-12T07:05:37.725087+00:00 | 2022-12-12T07:05:37.725125+00:00 | 726 | false | 1. I find the length of binary representation. For example, for 5 it will base-2 is "101" and length of this String is 3. \n2. Then by doing 1 << 3, I will have "1000" which 8 in base-10. \n3. Then I do 8 - 1 = 7, which is "111" in base-2.\n4. By applying XOR operation I will have "101" XOR "111" = "010", which is 2 in... | 5 | 0 | ['Java'] | 1 |
complement-of-base-10-integer | Python Simple Solution with Detail Explanation : O(log n) | python-simple-solution-with-detail-expla-qjml | There are 3 steps to be performed in this question:\n1. Convert Decimal (Base 10) number into to binary (Base 2)\n\t(We divide the number by 2 and write the rem | yashitanamdeo | NORMAL | 2022-01-04T15:55:55.214819+00:00 | 2022-01-12T14:53:33.183197+00:00 | 627 | false | There are 3 steps to be performed in this question:\n1. Convert Decimal (Base 10) number into to binary (Base 2)\n\t(We divide the number by `2` and write the remainders from bottom to top order)\n\t**Example:**\n\t | simple-java-sol-xor-by-poorvank-bf9z | \npublic class ComplementBase10 {\n\n public int bitwiseComplement(int N) {\n // Find number of bits in the given integer\n int bitCount = (int | poorvank | NORMAL | 2019-03-17T04:02:29.011313+00:00 | 2019-03-17T04:02:29.011371+00:00 | 881 | false | ```\npublic class ComplementBase10 {\n\n public int bitwiseComplement(int N) {\n // Find number of bits in the given integer\n int bitCount = (int) Math.floor(Math.log(N)/Math.log(2))+1;\n\n // XOR the given integer with math.pow(2,bitCount-1)\n return ((1 << bitCount) - 1) ^ N;\n }\n\... | 5 | 2 | [] | 1 |
complement-of-base-10-integer | 🚀🔥 Beating 100% 🚀 | ⚡ Ultimate Bitwise Complement Solution in Java! 💯✅ | beating-100-ultimate-bitwise-complement-a1eqy | 🧠 Intuition
The goal is to find the bitwise complement of a given number n.The bitwise complement flips all bits:
0 ➝ 1 ✅
1 ➝ 0 ❌
Instead of flipping each bit d | DanishaDS | NORMAL | 2025-02-15T04:36:32.735113+00:00 | 2025-02-15T04:36:32.735113+00:00 | 144 | false | 🧠 Intuition
The goal is to find the bitwise complement of a given number n.
The bitwise complement flips all bits:
0 ➝ 1 ✅
1 ➝ 0 ❌
Instead of flipping each bit directly, your approach reconstructs the complement by summing the powers of 2 corresponding to 0 bits. 🛠️
💡 Approach
1️⃣ Edge Cases:
If n == 0, return 1 ... | 4 | 0 | ['Java'] | 0 |
complement-of-base-10-integer | 💢☠💫Easiest👾Faster✅💯 Lesser🧠 🎯 C++✅Python3🐍✅Java✅C✅Python🐍✅C#✅💥🔥💫Explained☠💥🔥 Beats 100 | easiestfaster-lesser-cpython3javacpython-jq5r | Intuition\n\n\n\n\n\n Describe your first thoughts on how to solve this problem. \njavascript []\n//JavaScript Code\n/**\n * @param {number} n\n * @return {numb | Edwards310 | NORMAL | 2024-08-22T13:27:44.618342+00:00 | 2024-08-22T13:27:44.618369+00:00 | 568 | false | # Intuition\n\n\n\n\n\n# Code\n```\nclass Solution... | 4 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | BITWISE OPERATORS SOLUTIONS | bitwise-operators-solutions-by-2005115-xkga | PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT\n# CONNECT WITH ME\n### https://www.linkedin.com/in/pratay-nandy-9ba57b229/\n#### https://www.instagram.com/pratay_nand | 2005115 | NORMAL | 2023-10-13T05:16:29.125296+00:00 | 2023-10-13T05:16:29.125319+00:00 | 479 | false | # **PLEASE UPVOTE MY SOLUTION IF YOU LIKE IT**\n# **CONNECT WITH ME**\n### **[https://www.linkedin.com/in/pratay-nandy-9ba57b229/]()**\n#### **[https://www.instagram.com/pratay_nandy/]()**\n\n# Approach\nHere\'s a step-by-step explanation of the approach used in this code:\n\n1. Initialize `m` to the value of `n`. `m` ... | 4 | 0 | ['Bit Manipulation', 'C', 'C++'] | 0 |
complement-of-base-10-integer | Python Simple Solution Faster Than 99.49% | python-simple-solution-faster-than-9949-l0wb5 | \t\n\tclass Solution:\n def bitwiseComplement(self, n: int) -> int:\n s=bin(n).replace(\'0b\',\'\')\n t=\'\'\n for i in s:\n | pulkit_uppal | NORMAL | 2022-10-26T00:11:44.089132+00:00 | 2022-10-26T00:11:44.089174+00:00 | 616 | false | \t```\n\tclass Solution:\n def bitwiseComplement(self, n: int) -> int:\n s=bin(n).replace(\'0b\',\'\')\n t=\'\'\n for i in s:\n if i==\'0\':\n t+=\'1\'\n else:\n t+=\'0\'\n return int(t,2)\n``` | 4 | 0 | ['Python'] | 0 |
complement-of-base-10-integer | Easy Python Solution | O(1)T, O(1)S | Explanation | easy-python-solution-o1t-o1s-explanation-08y9 | The idea is to XOR the number with a mask. The mask should contain all 1\'s in it\'s rightmost bits. Here, the number of 1\'s is important. For example: for the | mohd_mustufa | NORMAL | 2022-01-04T18:08:31.833021+00:00 | 2022-01-04T18:08:31.833073+00:00 | 76 | false | The idea is to XOR the number with a mask. The mask should contain all 1\'s in it\'s rightmost bits. Here, the number of 1\'s is important. For example: for the number 5 (101), the mask should be 7 (111) which contains only 3 1\'s because the number 5 uses only 3 rightmost bits. Now when we XOR 5 and 7, we get 2:\n101 ... | 4 | 0 | ['Bitmask'] | 0 |
complement-of-base-10-integer | C++ using vector | c-using-vector-by-ayush_srivastava02-ybw6 | ```\n#define ull unsigned long long\nclass Solution {\npublic:\n vector binary(int n){\n vector bin;\n int temp=0;\n while(n>0){\n | ayush_srivastava02 | NORMAL | 2022-01-04T08:38:33.977673+00:00 | 2022-07-28T18:56:40.398615+00:00 | 214 | false | ```\n#define ull unsigned long long\nclass Solution {\npublic:\n vector<int> binary(int n){\n vector<int> bin;\n int temp=0;\n while(n>0){\n int rem = n%2;\n bin.push_back(rem);\n n/=2;\n }\n reverse(bin.begin(),bin.end());\n return bin;\n ... | 4 | 0 | ['C', 'C++'] | 0 |
complement-of-base-10-integer | C++ || Easy to Understand || Xor || Masking | c-easy-to-understand-xor-masking-by-luci-7fof | class Solution {\npublic:\n\n int bitwiseComplement(int n) {\n if(n==0){\n return 1;\n }\n int copy=n;\n int i=0;\n | luciferraturi | NORMAL | 2022-01-04T05:12:16.417699+00:00 | 2022-01-04T05:12:16.417769+00:00 | 165 | false | class Solution {\npublic:\n\n int bitwiseComplement(int n) {\n if(n==0){\n return 1;\n }\n int copy=n;\n int i=0;\n while(copy!=0){\n copy>>=1;\n n^=(1<<i);\n i++;\n }\n return n;\n }\n}; | 4 | 0 | ['Bit Manipulation', 'C', 'Bitmask', 'C++'] | 0 |
complement-of-base-10-integer | Simple Java solution | 0 ms beats 100% submissions | simple-java-solution-0-ms-beats-100-subm-srr5 | ```\n public int bitwiseComplement(int N) {\n if(N==0)\n return 1;\n int count = 0;\n int oldnum= N;\n \n //coun | yogidalal90 | NORMAL | 2020-10-05T12:42:37.926265+00:00 | 2020-10-05T14:05:07.258993+00:00 | 439 | false | ```\n public int bitwiseComplement(int N) {\n if(N==0)\n return 1;\n int count = 0;\n int oldnum= N;\n \n //count the num of digits in binary form for the given number\n while(N > 0){\n N >>=1;\n count++;\n }\n \n // generate ... | 4 | 3 | ['Bit Manipulation', 'Java'] | 1 |
complement-of-base-10-integer | Python simplest one-liner solution | python-simplest-one-liner-solution-by-ye-e01z | \nclass Solution:\n def bitwiseComplement(self, N: int) -> int:\n return (2**len(bin(N)[2:]))-1-N if N != 0 else 1\n\nLike it? please upvote... | yehudisk | NORMAL | 2020-10-05T08:43:52.488894+00:00 | 2020-10-05T08:43:52.488940+00:00 | 440 | false | ```\nclass Solution:\n def bitwiseComplement(self, N: int) -> int:\n return (2**len(bin(N)[2:]))-1-N if N != 0 else 1\n```\n**Like it? please upvote...** | 4 | 1 | ['Python3'] | 2 |
complement-of-base-10-integer | Java simple, short and easy solution, 0 ms - faster than 100% | java-simple-short-and-easy-solution-0-ms-a143 | \nclass Solution {\n int countDigits(int n) {\n if (n == 0)\n return 0;\n return 1 + countDigits(n>>1);\n }\n \n public int | yehudisk | NORMAL | 2020-10-05T08:35:11.656575+00:00 | 2020-10-05T08:35:11.656621+00:00 | 300 | false | ```\nclass Solution {\n int countDigits(int n) {\n if (n == 0)\n return 0;\n return 1 + countDigits(n>>1);\n }\n \n public int bitwiseComplement(int N) {\n if (N == 0) return 1;\n return (int)Math.pow(2, countDigits(N)) - 1 - N;\n }\n}\n```\n**Like it? please upvote... | 4 | 0 | ['Java'] | 0 |
complement-of-base-10-integer | C++ super-simple, short and easy solution, 0 ms, fatser than 100% | c-super-simple-short-and-easy-solution-0-ki3y | \nclass Solution {\npublic:\n int countDigits(int n) {\n if (n == 0)\n return 0;\n return 1 + countDigits(n>>1);\n }\n \n i | yehudisk | NORMAL | 2020-10-05T08:32:28.494161+00:00 | 2020-10-05T08:32:28.494192+00:00 | 239 | false | ```\nclass Solution {\npublic:\n int countDigits(int n) {\n if (n == 0)\n return 0;\n return 1 + countDigits(n>>1);\n }\n \n int bitwiseComplement(int N) {\n if (N == 0) return 1;\n return pow(2, countDigits(N)) - 1 - N;\n }\n};\n```\n**Like it? please upvote...** | 4 | 0 | ['C'] | 1 |
complement-of-base-10-integer | [Python, C++] LEGENDARY INNOVATIVE MAVERICK GENIUS SOLUTION; beats 100%, O(1) | python-c-legendary-innovative-maverick-g-rkwn | We need to XOR the number with a mask having as many 1s as the minimum no. of bits needed to represent the original number.\n\nFor instance:\n\n10 in binary wou | hipsuy | NORMAL | 2020-09-20T11:42:29.060128+00:00 | 2020-09-20T11:43:07.500278+00:00 | 294 | false | We need to XOR the number with a mask having as many 1s as the minimum no. of bits needed to represent the original number.\n\nFor instance:\n\n10 in binary would be -> 1010\nnotice that the no. of bits is "log2(10) + 1" \n\nThis holds true in general, for any decimal number x, number of bits needed to represent it in... | 4 | 0 | ['C', 'Python3'] | 0 |
complement-of-base-10-integer | [faster than 100.00%][c++][easy understanding] | faster-than-10000ceasy-understanding-by-zqzbj | \n//1.\nclass Solution {\npublic:\n int bitwiseComplement(int N) {\n if(N==0) \n\t\t return 1;\n int res=0,k=1;\n while(N){\n | rajat_gupta_ | NORMAL | 2020-09-15T18:14:14.516981+00:00 | 2020-09-15T18:14:14.517035+00:00 | 119 | false | ```\n//1.\nclass Solution {\npublic:\n int bitwiseComplement(int N) {\n if(N==0) \n\t\t return 1;\n int res=0,k=1;\n while(N){\n int d=N%2; //finding a bit\n res+=(d==1?0:1)*k; //complement of bit and complement\n N/=2; \n k*=2;\n }\n ... | 4 | 0 | ['C', 'C++'] | 0 |
complement-of-base-10-integer | c++ Solution using Bitwise ^ (Beats 100%) | c-solution-using-bitwise-beats-100-by-ra-0mx7 | XOR of any Number with a number having all bit 1 gives Binary complemented number.\nEg- 101 ^ 111 = 010 \n\n\nclass Solution {\npublic:\n int bitwiseCompleme | raven_13 | NORMAL | 2020-04-29T07:24:15.484825+00:00 | 2020-04-29T07:24:39.934420+00:00 | 182 | false | XOR of any Number with a number having all bit 1 gives Binary complemented number.\nEg- 101 ^ 111 = 010 \n\n```\nclass Solution {\npublic:\n int bitwiseComplement(int N) {\n \n if(N==0)\n return 1;\n int p=log2(N);\n\t\t\t\tlong num=pow(2,p+1)-1; // num has all bit 1;\n \n re... | 4 | 0 | [] | 0 |
complement-of-base-10-integer | PYTHON 5 lines, no need bit manipulation, just math | python-5-lines-no-need-bit-manipulation-0jkfv | \ndigit, num = 1, N\nwhile N > 1:\n\tdigit += 1\n\tN = N//2\nreturn 2**digit - 1 - num\n | kop_fyf | NORMAL | 2019-03-18T18:19:53.605860+00:00 | 2019-03-18T18:19:53.605902+00:00 | 623 | false | ```\ndigit, num = 1, N\nwhile N > 1:\n\tdigit += 1\n\tN = N//2\nreturn 2**digit - 1 - num\n``` | 4 | 0 | [] | 2 |
complement-of-base-10-integer | 💯Easy Solution | Mastering Complement of Base 10 | Beats 100% Solution | Bit Shifting 🚀 | mastering-complement-of-base-10-beats-10-fqv8 | IntuitionThe bitwise complement of a number n means flipping all its bits (0s become 1s and 1s become 0s). Instead of directly flipping the bits, we iterate thr | Shivam_Tyagii | NORMAL | 2025-02-15T13:34:54.021626+00:00 | 2025-02-16T03:57:27.966295+00:00 | 152 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
The bitwise complement of a number n means flipping all its bits (0s become 1s and 1s become 0s). Instead of directly flipping the bits, we iterate through each bit of n and construct the complement number step by step.
# Approach
<!-- Des... | 3 | 0 | ['Bit Manipulation', 'C++'] | 1 |
complement-of-base-10-integer | 100.00% | 3 line easy short | c++ code | math | O(log2(n)) | 10000-3-line-easy-short-c-code-math-olog-rlxd | Approach\n Describe your approach to solving the problem. \nJust find the just greater number than n which is power of 2 then subtact 1 from it.\nThen return xo | amanraox | NORMAL | 2024-08-22T15:56:49.306020+00:00 | 2024-08-22T15:56:49.306055+00:00 | 34 | false | # Approach\n<!-- Describe your approach to solving the problem. -->\nJust find the just greater number than n which is power of 2 then subtact 1 from it.\nThen return xor of both\n\n# Code\n```cpp []\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n unsigned int b=0,k=0;\n while(k<=n)b++,k=... | 3 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | 3 line easy short | c++ code | math | O(log2(n)) | 3-line-easy-short-c-code-math-olog2n-by-7jmys | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \njust find the just grea | amanraox | NORMAL | 2024-08-22T15:52:14.596454+00:00 | 2024-08-22T15:52:14.596480+00:00 | 27 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\njust find the just greater number than **n** which is power of 2 then subtact 1 from it.\nthen return xor of both\n\n# Code\n```cpp []\nclass Solution {\npublic:\n ... | 3 | 0 | ['Bit Manipulation', 'C++'] | 0 |
complement-of-base-10-integer | 🌟 Efficient Solution for "Complement of Base 10 Integer" | C++ | 0ms Beats 100.00% | O(log n) | efficient-solution-for-complement-of-bas-ymzq | \n\n---\n\n# Intuition\n\nTo find the complement of an integer, we need to flip all bits in its binary representation. For this, determine the number of bits re | user4612MW | NORMAL | 2024-08-22T04:40:30.353658+00:00 | 2024-08-22T04:41:07.730704+00:00 | 30 | false | ##\n\n---\n\n# **Intuition**\n\nTo find the complement of an integer, we need to flip all bits in its binary representation. For this, determine the number of bits required to represent the integer, create a bitmask with all bits set to 1 for this length, and then apply a bitwise XOR operation with the integer to get i... | 3 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | Easy and Simple C++ Approach ✅ | Beginner Friendly 🔥🔥🔥 | easy-and-simple-c-approach-beginner-frie-bqnn | Intuition\nTo find the bitwise complement of a number n, we need to flip all the bits of n. For example, if n is 5 (binary 101), its bitwise complement should b | AyushBansalCodes | NORMAL | 2024-08-03T07:19:43.063390+00:00 | 2024-08-03T07:19:43.063441+00:00 | 184 | false | # Intuition\nTo find the bitwise complement of a number `n`, we need to flip all the bits of `n`. For example, if `n` is `5` (binary `101`), its bitwise complement should be `2` (binary `010`).\n\n# Approach\n1. If `n` is `0`, directly return `1` because the complement of `0` is `1`.\n1. Create a mask that has the same... | 3 | 0 | ['Bit Manipulation', 'C++'] | 0 |
complement-of-base-10-integer | C++ | 0 ms | Beats 100% | c-0-ms-beats-100-by-djwhocodes_5-o4ea | Intuition\n Describe your first thoughts on how to solve this problem. \nThe code handles the base case where num is 0 by returning 1, as the bitwise complement | DJwhoCODES_5 | NORMAL | 2024-03-17T05:18:37.110942+00:00 | 2024-03-17T05:18:37.110965+00:00 | 222 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe code handles the base case where num is 0 by returning 1, as the bitwise complement of 0 is 1.\n\nIt iterates through each bit of num to compute its bitwise complement by flipping each bit individually.\n\nIt does this by extracting e... | 3 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | Simple Python Solution || O(logN) | simple-python-solution-ologn-by-dodda_sa-bq3y | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Dodda_Sai_Sachin | NORMAL | 2024-02-27T07:11:57.386786+00:00 | 2024-02-27T07:11:57.386844+00:00 | 519 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['Python3'] | 0 |
complement-of-base-10-integer | Easy Solution || C++ | easy-solution-c-by-japneet001-tmb8 | Intuition\nThe problem requires finding the bitwise complement of a given integer. Bitwise complement is the operation of flipping the bits of a binary represen | Japneet001 | NORMAL | 2024-02-01T04:08:04.797488+00:00 | 2024-02-01T04:08:04.797519+00:00 | 107 | false | # Intuition\nThe problem requires finding the bitwise complement of a given integer. Bitwise complement is the operation of flipping the bits of a binary representation of a number (changing 0s to 1s and vice versa).\n\n# Approach\nThe provided code uses a bitwise operation to iterate through the bits of the input numb... | 3 | 0 | ['Bit Manipulation', 'C++'] | 0 |
complement-of-base-10-integer | 100% Beats Solution || Using Bit Manipulation || Step by Step Explain || Easy to Understand || | 100-beats-solution-using-bit-manipulatio-h2f7 | Abhiraj Pratap Singh\n\n---\n\n# UPVOTE IT....\n\n---\n\n# Intuition\n Describe your first thoughts on how to solve this problem. \n\n- The goal of the function | abhirajpratapsingh | NORMAL | 2023-09-27T17:35:41.072440+00:00 | 2024-01-16T16:06:48.836779+00:00 | 7 | false | # Abhiraj Pratap Singh\n\n---\n\n# UPVOTE IT....\n\n---\n\n# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n- The goal of the function is to find the bitwise complement of a given integer n. The bitwise complement of a binary number is obtained by flipping each bit (changing 0s to 1s ... | 3 | 0 | ['Bit Manipulation', 'C++'] | 0 |
complement-of-base-10-integer | Python 2 Approach 97.96% beats | python-2-approach-9796-beats-by-vvivekya-zofy | If you got help from this,... Plz Upvote .. it encourage me\n# EXAMPLE , \n # N = 101 (5)\n # MASK WILL BE 111 (7)\n # COMPLEMENT MASK - | vvivekyadav | NORMAL | 2023-09-22T13:00:10.518194+00:00 | 2023-09-23T05:53:42.464679+00:00 | 182 | false | **If you got help from this,... Plz Upvote .. it encourage me**\n# EXAMPLE , \n # N = 101 (5)\n # MASK WILL BE 111 (7)\n # COMPLEMENT MASK - N = 7-5 = 2 (010)\n# Code\n```\n<!-- 1ST APPROACH -->\nclass Solution:\n def bitwiseComplement(self, n: int) -> int:\n \n if n == 0:\n ... | 3 | 0 | ['Bit Manipulation', 'Python', 'Python3'] | 1 |
complement-of-base-10-integer | C++ Solution || Runtime 0ms || Beats 100% || | c-solution-runtime-0ms-beats-100-by-asad-7lfa | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Asad_Sarwar | NORMAL | 2023-02-18T21:02:40.618487+00:00 | 2023-02-18T21:02:40.618529+00:00 | 726 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['C++'] | 1 |
complement-of-base-10-integer | Easy C++ solution || Beats 100% || 0ms solution || O(log(n)) solution | easy-c-solution-beats-100-0ms-solution-o-tx8t | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | umangrathi110 | NORMAL | 2022-10-19T13:24:23.971348+00:00 | 2022-10-19T13:24:23.971388+00:00 | 604 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity: O(log(N))\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity: O(1)\n<!-- Add your space complexity here, e... | 3 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | 3ms || C++ || EASY || SIMPLE || TIME O(position of mst) || SPACE O(1) | 3ms-c-easy-simple-time-oposition-of-mst-cuh8f | \nclass Solution {\npublic:\n int bitwiseComplement(int &n) {\n if(n==0)return 1;\n int ans = 0;\n int i = 0;\n while(n){\n | abhay_12345 | NORMAL | 2022-09-08T07:38:58.188371+00:00 | 2022-09-08T07:38:58.188409+00:00 | 532 | false | ```\nclass Solution {\npublic:\n int bitwiseComplement(int &n) {\n if(n==0)return 1;\n int ans = 0;\n int i = 0;\n while(n){\n ans += ((n%2)^1)<<i;\n i++;\n n = n >> 1;\n }\n return ans;\n }\n};\n``` | 3 | 0 | ['Bit Manipulation', 'C', 'C++'] | 1 |
complement-of-base-10-integer | (newest approach) c++ 100% 0 ms without converting number into binary | newest-approach-c-100-0-ms-without-conve-nnst | \nclass Solution {\npublic:\n int power_floor( int n)\n {\n \n if((n!=0) &&!(n & (n-1)))\n {\n return n;\n }\n | avani_baheti_59_70 | NORMAL | 2022-05-27T07:26:35.006305+00:00 | 2023-02-28T10:18:22.542618+00:00 | 135 | false | ```\nclass Solution {\npublic:\n int power_floor( int n)\n {\n \n if((n!=0) &&!(n & (n-1)))\n {\n return n;\n }\n int count=0;\n while(n!=0)\n {\n n>>=1;\n count+=1;\n }\n return 1<<count;\n }\n int bitwiseComplem... | 3 | 0 | ['Bit Manipulation'] | 0 |
complement-of-base-10-integer | C++ | 0 ms | One Line Solution using XOR | c-0-ms-one-line-solution-using-xor-by-mu-khci | \nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n if(n==0) return 1;\n \n return (int)(pow(2,(int)log2(n) + 1) - 1) ^ n ;\n | mukulnayak | NORMAL | 2022-04-10T07:04:20.431161+00:00 | 2022-04-10T07:05:57.405300+00:00 | 176 | false | ```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n if(n==0) return 1;\n \n return (int)(pow(2,(int)log2(n) + 1) - 1) ^ n ;\n }\n};\n```\n\nExplanation: we have to make ***0101*** from 1010.\nwe observe, 1010 ^ *0101* = 1111\nHence,\nXOR with 1010 on both sides we get (we know t... | 3 | 0 | ['Bit Manipulation', 'C'] | 0 |
complement-of-base-10-integer | Easy Solution, Faster than 100%, Observation Based | easy-solution-faster-than-100-observatio-chyk | \nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n if(n==0)\n return 1;\n int i=1;\n while(i<=32){\n i | vasu_2006 | NORMAL | 2022-02-14T14:36:33.352584+00:00 | 2022-02-14T14:36:33.352626+00:00 | 48 | false | ```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n if(n==0)\n return 1;\n int i=1;\n while(i<=32){\n if(n-pow(2,i)<0)\n break;\n i++;\n }\n return pow(2,i)-n-1;\n }\n};\n``` | 3 | 0 | ['Bit Manipulation'] | 0 |
complement-of-base-10-integer | Python 3 (50ms) | Beginner Naive Approach | String, Join & Replace | python-3-50ms-beginner-naive-approach-st-njt9 | \nclass Solution:\n def bitwiseComplement(self, n: int) -> int:\n t=(str(bin(n).replace("0b", "")))\n a="".join("1" if x==\'0\' else \'0\' for | MrShobhit | NORMAL | 2022-01-19T09:18:41.913334+00:00 | 2022-01-19T09:18:41.913370+00:00 | 207 | false | ```\nclass Solution:\n def bitwiseComplement(self, n: int) -> int:\n t=(str(bin(n).replace("0b", "")))\n a="".join("1" if x==\'0\' else \'0\' for x in t)\n return int(a,2)\n``` | 3 | 0 | ['String', 'Python'] | 2 |
complement-of-base-10-integer | Java Soln if you don't know Bit Manipulation(Just like me) | java-soln-if-you-dont-know-bit-manipulat-50fa | \nclass Solution {\n public int bitwiseComplement(int n) {\n \n if(n ==0) return 1 ; //If we have 0, the complement will be 1\n \n | vbhv_sxn | NORMAL | 2022-01-04T17:41:05.731708+00:00 | 2022-01-05T18:23:41.081074+00:00 | 85 | false | ```\nclass Solution {\n public int bitwiseComplement(int n) {\n \n if(n ==0) return 1 ; //If we have 0, the complement will be 1\n \n // next = Just greater no. which is Power of 2\n //Using the formulae\'loga b = log10 b / log10 a\' we can calculate log to any base\n int ne... | 3 | 0 | ['Java'] | 0 |
complement-of-base-10-integer | Faster than 100% | faster-than-100-by-vasu_2006-bw1c | class Solution {\npublic:\n int bitwiseComplement(int n) {\n if(n==0)\n return 1;\n int i=1;\n while(i<=32){\n if( | vasu_2006 | NORMAL | 2022-01-04T13:51:18.346752+00:00 | 2022-01-04T13:51:18.346789+00:00 | 33 | false | class Solution {\npublic:\n int bitwiseComplement(int n) {\n if(n==0)\n return 1;\n int i=1;\n while(i<=32){\n if(n-pow(2,i)<0)\n break;\n i++;\n }\n return pow(2,i)-n-1;\n }\n}; | 3 | 0 | [] | 0 |
complement-of-base-10-integer | Easy to understand JavaScript solution | easy-to-understand-javascript-solution-b-esav | How to find an answer without the knowledge of bitwise operators?\n\nThe hint lies here: to find the complement number, we need to find the 111..111 number of N | besLisbeth | NORMAL | 2022-01-04T09:34:12.975469+00:00 | 2022-01-04T09:34:12.975499+00:00 | 215 | false | How to find an answer without the knowledge of bitwise operators?\n\nThe hint lies here: to find the complement number, we need to find the 111..111 number of N length, where N is the length of the input number in binary representation. That\'s our sum of `n` and complement number. Then convert it to decimal and subtra... | 3 | 0 | ['TypeScript', 'JavaScript'] | 0 |
complement-of-base-10-integer | The sum of a number and it's complement is all 111's. | the-sum-of-a-number-and-its-complement-i-jeir | \'\'\'\nclass Solution {\npublic:\n int bitwiseComplement(int N)\n {\n if(N==0)return 1; \n int sum = (int)pow(2,1+(int)log2(N))-1;\n | varchirag | NORMAL | 2022-01-04T05:09:55.114961+00:00 | 2022-01-04T05:09:55.115014+00:00 | 91 | false | \'\'\'\nclass Solution {\npublic:\n int bitwiseComplement(int N)\n {\n if(N==0)return 1; \n int sum = (int)pow(2,1+(int)log2(N))-1;\n return sum-N;\n }\n}; | 3 | 0 | ['Math', 'C', 'C++'] | 1 |
complement-of-base-10-integer | ✔️ [Python3] SHIFT AND XOR (ᕗ ͠° ਊ ͠° )ᕗ, Explained | python3-shift-and-xor-fo-deg-uu-deg-fo-e-f5z1 | We create a mask=1 and shift its bit to the left until it gets greater than n. Example: n=101 => mask=1000. After, we substruct 1 from the mask which gives us a | artod | NORMAL | 2022-01-04T00:30:39.815823+00:00 | 2022-01-04T00:30:39.815854+00:00 | 277 | false | We create a mask=`1` and shift its bit to the left until it gets greater than `n`. Example: `n=101` => `mask=1000`. After, we substruct `1` from the mask which gives us all `1`\'s: `mask=111`. And the last step, we simply return xor between `n` and our mask: `n ^ mask` = `101 ^ 111` = `010`.\n\nTime: **O(1)** - since ... | 3 | 3 | ['Python3'] | 1 |
complement-of-base-10-integer | Go solution faster than 100% bit manuplation | go-solution-faster-than-100-bit-manuplat-1si2 | \nfunc bitwiseComplement(N int) int {\n\tif N == 0 {\n\t\treturn 1\n\t}\n\tres := 0\n\tg := 1\n\tfor N > 0 {\n\t\tif N&1 == 0 {\n\t\t\tres += g\n\t\t}\n\t\tg << | nathannaveen | NORMAL | 2021-02-10T16:48:53.033527+00:00 | 2021-02-10T16:48:53.033555+00:00 | 194 | false | ```\nfunc bitwiseComplement(N int) int {\n\tif N == 0 {\n\t\treturn 1\n\t}\n\tres := 0\n\tg := 1\n\tfor N > 0 {\n\t\tif N&1 == 0 {\n\t\t\tres += g\n\t\t}\n\t\tg <<= 1\n\t\tN >>= 1\n\t}\n\treturn res\n}\n``` | 3 | 0 | ['Bit Manipulation', 'Go'] | 0 |
complement-of-base-10-integer | [C++] Math based 3 line no loop | c-math-based-3-line-no-loop-by-sameerp98-uaam | Math approach\nThe sum of a number and it\'s complement is all 111\'s. For example :\n4 + 3 = 7 (100 + 011= 111)\nNow if we want to find complement we can simpl | sameerp98 | NORMAL | 2020-10-05T18:56:28.374514+00:00 | 2022-01-20T20:46:24.811957+00:00 | 256 | false | # Math approach\nThe sum of a number and it\'s complement is all 111\'s. For example :\n4 + 3 = 7 (100 + 011= 111)\nNow if we want to find complement we can simply subract our number from a number which has all 1\'s from Most significant bit (MSB) to Least significant bit (LSB).\nTo find the most significant bit of a b... | 3 | 0 | ['Math', 'C', 'C++'] | 0 |
complement-of-base-10-integer | C++ One Line Solution 100% faster 0ms | c-one-line-solution-100-faster-0ms-by-de-8nue | \nclass Solution {\npublic:\n int bitwiseComplement(int N) \n {\n return N == 0 ? 1 : N ^ (int)(pow(2, floor(log2(N)) + 1) - 1);\n }\n};\n | debbiealter | NORMAL | 2020-10-05T08:59:59.904898+00:00 | 2020-10-05T10:32:02.258255+00:00 | 494 | false | ```\nclass Solution {\npublic:\n int bitwiseComplement(int N) \n {\n return N == 0 ? 1 : N ^ (int)(pow(2, floor(log2(N)) + 1) - 1);\n }\n};\n``` | 3 | 1 | ['C', 'C++'] | 0 |
complement-of-base-10-integer | Python super-simple short & clean solution | python-super-simple-short-clean-solution-dnmt | \nclass Solution:\n def bitwiseComplement(self, N: int) -> int:\n if N==0:\n return 1\n \n def countDigits(n):\n i | yehudisk | NORMAL | 2020-10-05T08:27:56.134752+00:00 | 2020-10-05T08:27:56.134788+00:00 | 303 | false | ```\nclass Solution:\n def bitwiseComplement(self, N: int) -> int:\n if N==0:\n return 1\n \n def countDigits(n):\n if n == 0:\n return 0\n return 1 + countDigits(n>>1)\n \n return (2**countDigits(N))-1-N\n```\n**Like it? please upvot... | 3 | 0 | ['Python3'] | 1 |
complement-of-base-10-integer | c++ solution ,beats 100%(0ms, 5MB) | c-solution-beats-1000ms-5mb-by-goyalansh-anbd | class Solution {\npublic:\n int bitwiseComplement(int n) {\n \n if(n==0)\n return 1;\n int countbits = floor(log2(n)) + 1;\n | goyalanshul809 | NORMAL | 2020-08-17T05:56:24.935367+00:00 | 2020-08-17T05:56:24.935400+00:00 | 82 | false | class Solution {\npublic:\n int bitwiseComplement(int n) {\n \n if(n==0)\n return 1;\n int countbits = floor(log2(n)) + 1;\n long long int temp = pow(2,countbits);\n temp--;\n long long int ans = n^temp;\n return ans;\n }\n}; | 3 | 0 | [] | 0 |
complement-of-base-10-integer | Java 0ms, No bitwise, Easy to understand | java-0ms-no-bitwise-easy-to-understand-b-7md9 | ```\npublic int findComplement(int num) {\n\n\tint i = 0;\n\tint res = 0;\n\t\n\twhile (num > 0) {\n\t\n\t\tif (num % 2 == 0)\n\t\t\tres += Math.pow(2, i);\n\t\ | nive00 | NORMAL | 2020-05-05T04:52:42.066576+00:00 | 2020-05-05T04:54:56.727056+00:00 | 322 | false | ```\npublic int findComplement(int num) {\n\n\tint i = 0;\n\tint res = 0;\n\t\n\twhile (num > 0) {\n\t\n\t\tif (num % 2 == 0)\n\t\t\tres += Math.pow(2, i);\n\t\ti++;\n\t\tnum /= 2;\n\t}\n\treturn res;\n}\n\t | 3 | 1 | [] | 0 |
complement-of-base-10-integer | C++ : Beats 100% time and 100% space | c-beats-100-time-and-100-space-by-metacr-nqgk | My approach invloves working from the rightmost bit of the given number.\nIf the rightmost bit is 1, we add 0. Otherwise, we add the power of two correspoding t | metacresol | NORMAL | 2020-05-04T11:30:36.809481+00:00 | 2020-05-04T11:31:17.216126+00:00 | 675 | false | My approach invloves working from the rightmost bit of the given number.\nIf the rightmost bit is 1, we add 0. Otherwise, we add the power of two correspoding to the bit, which we keep track here using the variable \'pow\'. After each check we rightshift the number.\n```\nclass Solution {\npublic:\n int bitwiseCompl... | 3 | 0 | ['C++'] | 1 |
complement-of-base-10-integer | Javascript solution | javascript-solution-by-mmartire-15q5 | ```\n/*\n * @param {number} N\n * @return {number}\n /\nvar bitwiseComplement = function(N) {\n return ((1 << (N.toString(2).length)) - 1) ^ N;\n}; | mmartire | NORMAL | 2019-09-07T01:55:40.798371+00:00 | 2019-09-07T01:55:40.798430+00:00 | 420 | false | ```\n/**\n * @param {number} N\n * @return {number}\n */\nvar bitwiseComplement = function(N) {\n return ((1 << (N.toString(2).length)) - 1) ^ N;\n}; | 3 | 0 | ['JavaScript'] | 1 |
complement-of-base-10-integer | Java || Beats 100% || Bit Manipulation | java-beats-100-bit-manipulation-by-shikh-aoyh | Complexity
Time complexity: O(log N)
Space complexity:O(1)
Code | shikhargupta0645 | NORMAL | 2025-02-11T06:19:13.675462+00:00 | 2025-02-11T06:19:13.675462+00:00 | 75 | false |
# Complexity
- Time complexity: O(log N)
<!-- Add your time complexity here, e.g. $$O(n)$$ -->
- Space complexity:O(1)
<!-- Add your space complexity here, e.g. $$O(n)$$ -->
# Code
```java []
class Solution {
public int bitwiseComplement(int n) {
if(n==0){
return 1;
}
int ans... | 2 | 0 | ['Java'] | 0 |
complement-of-base-10-integer | Easily Understandable solution || 100% T.C ||CPP | easily-understandable-solution-100-tc-cp-lv49 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Ganesh_ag10 | NORMAL | 2024-07-03T15:23:28.817315+00:00 | 2024-07-03T15:23:28.817346+00:00 | 104 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | Finding the bitwise complement | finding-the-bitwise-complement-by-theath-swnq | Intuition\n Describe your first thoughts on how to solve this problem. \nThe problem requires finding the bitwise complement of a given integer. One way to appr | theatharvv | NORMAL | 2024-04-17T14:40:43.037151+00:00 | 2024-04-17T14:40:43.037185+00:00 | 13 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\nThe problem requires finding the bitwise complement of a given integer. One way to approach this is to iterate through each bit of the integer, flip it, and construct the complement.\n\n# Approach\n<!-- Describe your approach to solving t... | 2 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | easy solution | beats 100% | easy-solution-beats-100-by-leet1101-w7h3 | Approach\n1. If the input number is 0, return 1 as the bitwise complement of 0 is 1.\n2. Determine the number of bits required to represent the input number by | leet1101 | NORMAL | 2024-04-12T15:47:51.038989+00:00 | 2024-04-12T19:08:37.461518+00:00 | 380 | false | # Approach\n1. If the input number is 0, return 1 as the bitwise complement of 0 is 1.\n2. Determine the number of bits required to represent the input number by counting the number of shifts needed until the number becomes 0.\n3. Create a mask with all bits set to 1 using the formula \\(2^i - 1\\), where \\(i\\) is th... | 2 | 0 | ['Bit Manipulation', 'C++'] | 1 |
complement-of-base-10-integer | Code to find complement of Base 10 integer | code-to-find-complement-of-base-10-integ-uj9o | Intuition\nIn this question we have to find the complement of the given number. So we will simply do bitwise complement of the given number.\nFor eg., let x=100 | Parul_Awasthi | NORMAL | 2024-03-16T06:17:18.555211+00:00 | 2024-03-16T06:17:18.555244+00:00 | 23 | false | # Intuition\nIn this question we have to find the complement of the given number. So we will simply do bitwise complement of the given number.\nFor eg., let x=10011 so it\'s complement would be 01100.\n\n# Approach\nWe wil initialize few variables in the starting that are mul and sum for storing the power of 2\'s and t... | 2 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | Simple Java Code || Beats 100% | simple-java-code-beats-100-by-saurabh_mi-wesy | Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)\n# Code\n\nclass Solution {\n public int bitwiseComplement(int n) {\n int allOne = 1;\ | Saurabh_Mishra06 | NORMAL | 2024-02-22T10:05:20.215016+00:00 | 2024-02-22T10:05:20.215050+00:00 | 253 | false | # Complexity\n- Time complexity:\nO(n)\n- Space complexity:\nO(1)\n# Code\n```\nclass Solution {\n public int bitwiseComplement(int n) {\n int allOne = 1;\n while(allOne < n){\n allOne = allOne*2+1;\n }\n return allOne - n;\n }\n}\n``` | 2 | 0 | ['Java'] | 0 |
complement-of-base-10-integer | 100% efficient and easy | 100-efficient-and-easy-by-desalgentamira-5sdw | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | DesalgenTamirat | NORMAL | 2024-01-13T18:34:38.884222+00:00 | 2024-01-13T18:34:38.884244+00:00 | 1,180 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Python3'] | 1 |
complement-of-base-10-integer | 🚀Beats 100%🔥🤩||Super Clean✅||Easy code with explanation.. | beats-100super-cleaneasy-code-with-expla-kkfa | Intuition & Approach\n Describe your approach to solving the problem. \n1. The code constructs a bitmask that has all bits set up to the most significant bit of | Saurav_Kr_Singh | NORMAL | 2024-01-02T15:34:02.729469+00:00 | 2024-01-02T15:34:02.729492+00:00 | 49 | false | # Intuition & Approach\n<!-- Describe your approach to solving the problem. -->\n1. The code constructs a bitmask that has all bits set up to the most significant bit of n by shifting and OR operations.\n2. It then uses bitwise NOT (~) on n to flip all bits.\n3. Finally, it performs a bitwise AND with the bitmask to ge... | 2 | 0 | ['Bit Manipulation', 'Bitmask', 'C++'] | 0 |
complement-of-base-10-integer | Using Bit Masking || Beats 100% || JAVA | using-bit-masking-beats-100-java-by-ramu-b42l | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ramuyadav90597 | NORMAL | 2023-12-25T14:31:37.592888+00:00 | 2023-12-25T14:31:37.592924+00:00 | 327 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
complement-of-base-10-integer | Easy intuition, approach and code in C++ | easy-intuition-approach-and-code-in-c-by-uu2x | Intuition\nThe basic logic is that we will make a mask and make it bits 1 , equal to the number of bits in the given number and then take bitwise AND with the c | arpit123boss | NORMAL | 2023-10-31T17:33:48.918683+00:00 | 2023-10-31T17:33:48.918714+00:00 | 1,174 | false | # Intuition\nThe basic logic is that we will make a mask and make it bits 1 , equal to the number of bits in the given number and then take bitwise AND with the compliment of the given number. \n\n# Approach\n- First of all, we will make a mask variable and initialize it with 0.\n- Then store the given number to anothe... | 2 | 0 | ['C++'] | 1 |
complement-of-base-10-integer | Best Java Solution || Beats 100% | best-java-solution-beats-100-by-ravikuma-iz0f | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | ravikumar50 | NORMAL | 2023-09-24T07:55:01.140457+00:00 | 2023-09-24T07:55:01.140482+00:00 | 285 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Java'] | 0 |
complement-of-base-10-integer | C++ | Bit Manipulation | O(N) | c-bit-manipulation-on-by-omjadhav007-rzmh | Intuition\n\n\n# Approach\nBit Manipulation\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n\nclass Solution {\npublic:\n int | omjadhav007 | NORMAL | 2023-09-16T18:11:19.786438+00:00 | 2023-09-16T18:11:19.786464+00:00 | 0 | false | # Intuition\n\n\n# Approach\nBit Manipulation\n\n# Complexity\n- Time complexity: O(n)\n\n- Space complexity: O(1)\n\n# Code\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n int x=0,m=1;\n if(n==0) return 1;\n while(n){\n if(!(n&1)) x+=m;\n m*=2;\n ... | 2 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | Beginner Friendly Python || Beats 91% | beginner-friendly-python-beats-91-by-ana-azwx | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \nLet\'s assume a number, | anakamboj | NORMAL | 2023-09-15T17:38:07.242830+00:00 | 2023-09-15T17:38:07.242856+00:00 | 173 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\nLet\'s assume a number, say 10. \n\n```\nBinary version = 1010\nComplement = 0101\nInteger compelement = 5\n```\n\n```\nmask is initially 0,\nTill m = 10 (1010) is not... | 2 | 0 | ['Python3'] | 1 |
complement-of-base-10-integer | Java Code with Explanation | java-code-with-explanation-by-athravmeht-nts3 | Code\n\nclass Solution {\n public int bitwiseComplement(int n) {\n // edge Case \n if(n == 0) return 1;\n\n // ~n or can be said as NOT | athravmehta06 | NORMAL | 2023-07-17T06:55:53.392072+00:00 | 2023-07-17T06:55:53.392092+00:00 | 200 | false | # Code\n```\nclass Solution {\n public int bitwiseComplement(int n) {\n // edge Case \n if(n == 0) return 1;\n\n // ~n or can be said as NOT of n reverses all the bits of the number\n // we will write shift until our no. is not equal to zero so that we can find how many bits have our answ... | 2 | 0 | ['Bit Manipulation', 'Java'] | 0 |
complement-of-base-10-integer | easiest solution begineer friendly c++ | easiest-solution-begineer-friendly-c-by-691nm | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | gouravsaroha | NORMAL | 2023-04-27T03:37:09.338070+00:00 | 2023-04-27T03:37:09.338107+00:00 | 237 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['C++'] | 1 |
complement-of-base-10-integer | Simple Soluion | simple-soluion-by-arunarunarun7354-rl77 | \n\n# Code\n\nclass Solution {\n public int bitwiseComplement(int n) {\n String bin = Integer.toBinaryString(n);\n String res = "";\n fo | arunarunarun7354 | NORMAL | 2023-04-03T23:01:22.117875+00:00 | 2023-04-03T23:01:22.117927+00:00 | 539 | false | \n\n# Code\n```\nclass Solution {\n public int bitwiseComplement(int n) {\n String bin = Integer.toBinaryString(n);\n String res = "";\n for(char c :bin.toCharArray())\n {\n if( c == \'1\')\n res += "0";\n else\n res += "1";\n }\n... | 2 | 0 | ['Java'] | 0 |
complement-of-base-10-integer | ✅ [C++] EASY and Shorter Solution || Intuitive Approach || 100% Accepted | c-easy-and-shorter-solution-intuitive-ap-l4n3 | Intuition\n1. Straight formula based question.\n\n# Approach\n1. Count the number of digits in the number.\n2. Use the formula :- one complement of x is 2^n-x-1 | akh1465 | NORMAL | 2023-03-17T06:22:21.470253+00:00 | 2023-03-17T06:22:21.470280+00:00 | 1,106 | false | # Intuition\n1. Straight formula based question.\n\n# Approach\n1. Count the number of digits in the number.\n2. Use the formula :- one complement of x is 2^n-x-1\n\n# Complexity\n- Time complexity:\nO(N)\n\n- Space complexity:\nO(1)\n\n# Code\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n\n ... | 2 | 0 | ['C++'] | 2 |
complement-of-base-10-integer | C++ solution faster than 100% of the solutions | c-solution-faster-than-100-of-the-soluti-8yqk | \n# Code\n\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n int m = n ; \n int mask = 0 ; \n \n while(m==0){\n | ackerman217 | NORMAL | 2023-01-04T11:03:59.716283+00:00 | 2023-01-04T11:03:59.716324+00:00 | 1,204 | false | \n# Code\n```\nclass Solution {\npublic:\n int bitwiseComplement(int n) {\n int m = n ; \n int mask = 0 ; \n \n while(m==0){\n return 1 ; \n }\n while(m!=0){\n mask = ( mask << 1 ) | 1 ;\n m = m >> 1 ; \n }\n int ans = (~n) & ma... | 2 | 0 | ['C++'] | 1 |
complement-of-base-10-integer | C++ easy solution beats 100% | c-easy-solution-beats-100-by-ackerman217-saz4 | Intuition\nMake a mask \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time complexity:\n Add your time complexity here, e.g | ackerman217 | NORMAL | 2022-12-30T20:58:01.592485+00:00 | 2022-12-30T20:58:01.592517+00:00 | 631 | false | # Intuition\nMake a mask \n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution {\npublic:\n int bi... | 2 | 0 | ['C++'] | 0 |
complement-of-base-10-integer | Easy Java Solution Using Bit Manipulation ( Beats : 100 %) | easy-java-solution-using-bit-manipulatio-ojbv | \n# Approach\n Describe your approach to solving the problem. \nBit Manipulation\n\nHope this helps.\nDo Upvote if you like it !!\n\nThanks :)\n\n# Code\n\nclas | aishi535 | NORMAL | 2022-12-28T04:09:33.289839+00:00 | 2023-01-06T02:21:42.331950+00:00 | 616 | false | \n# Approach\n<!-- Describe your approach to solving the problem. -->\nBit Manipulation\n\nHope this helps.\nDo Upvote if you like it !!\n\nThanks :)\n\n# Code\n```\nclass Solution {\n public int bitwiseComplement(int num) {\n if(num==0){\n return 1;\n }\n int c=0;\n c=(int)(Ma... | 2 | 0 | ['Bit Manipulation', 'Java'] | 0 |
complement-of-base-10-integer | Python ✅✅✅ || 5-line Faster than 98.91% || Memory Beats 96.46% | python-5-line-faster-than-9891-memory-be-w7vy | Code\n\nclass Solution:\n def bitwiseComplement(self, n: int) -> int:\n b = bin(n)[2:]\n b = b.replace(\'1\', \'_\')\n b = b.replace(\'0 | a-fr0stbite | NORMAL | 2022-12-16T05:12:12.643542+00:00 | 2022-12-16T05:12:12.643578+00:00 | 179 | false | # Code\n```\nclass Solution:\n def bitwiseComplement(self, n: int) -> int:\n b = bin(n)[2:]\n b = b.replace(\'1\', \'_\')\n b = b.replace(\'0\', \'1\')\n b = b.replace(\'_\', \'0\')\n return int(b, 2)\n```\n of a given DataFrame.\n\n# Approach\nTo solve this problem, we can use the Pandas l | vaish_1929 | NORMAL | 2023-10-07T08:20:45.627372+00:00 | 2023-10-07T08:20:45.627400+00:00 | 8,848 | false | # Intuition\nThe problem asks us to find the size (number of rows and columns) of a given DataFrame.\n\n# Approach\nTo solve this problem, we can use the Pandas library in Python. We can directly use the `.shape` attribute of the DataFrame to get the number of rows and columns. We return the result as an array containi... | 33 | 0 | ['Pandas'] | 0 |
get-the-size-of-a-dataframe | Pandas 1-Line | Elegant & Short | And more Pandas solutions ✅ | pandas-1-line-elegant-short-and-more-pan-tztu | Complexity\n- Time complexity: O(1)\n- Space complexity: O(1)\n\n# Code\nPython []\ndef getDataframeSize(players: pd.DataFrame) -> list[int]:\n return list(p | Kyrylo-Ktl | NORMAL | 2023-10-06T09:06:45.451797+00:00 | 2023-10-06T09:06:45.451818+00:00 | 2,077 | false | # Complexity\n- Time complexity: $$O(1)$$\n- Space complexity: $$O(1)$$\n\n# Code\n```Python []\ndef getDataframeSize(players: pd.DataFrame) -> list[int]:\n return list(players.shape)\n```\n\n# Important!\n###### If you like the solution or find it useful, feel free to **upvote** for it, it will support me in creati... | 16 | 0 | ['Python', 'Python3', 'Pandas'] | 0 |
get-the-size-of-a-dataframe | ✅Easy And Simple Solution With EXPLANATION✅ | easy-and-simple-solution-with-explanatio-vevr | \n\n### Explanation\n\nThis code defines a function named getDataframeSize that takes a pandas DataFrame as input and returns its dimensions (number of rows and | deleted_user | NORMAL | 2024-06-07T07:29:52.795139+00:00 | 2024-06-07T07:29:52.795167+00:00 | 3,922 | false | \n\n### Explanation\n\nThis code defines a function named `getDataframeSize` that takes a pandas DataFrame as input and returns its dimensions (number of rows and columns) as a list.\n\n#### Step-by-step Explanation:\n\n1. **Import pandas**:\n ```python\n import pandas as pd\n ```\n - This line imports the ... | 11 | 0 | ['Pandas'] | 2 |
get-the-size-of-a-dataframe | 1 line code || Simple code|| (use shape property) | 1-line-code-simple-code-use-shape-proper-6l4j | If you got help from this,... Plz Upvote .. it encourage me\n# Code\n\nimport pandas as pd\n\ndef getDataframeSize(players: pd.DataFrame) -> List[int]:\n ret | vvivekyadav | NORMAL | 2023-10-06T00:43:39.188977+00:00 | 2023-10-06T16:08:48.770081+00:00 | 887 | false | **If you got help from this,... Plz Upvote .. it encourage me**\n# Code\n```\nimport pandas as pd\n\ndef getDataframeSize(players: pd.DataFrame) -> List[int]:\n return [players.shape[0], players.shape[1]]\n # OR\n return list(players.shape)\n``` | 8 | 0 | ['Python', 'Python3', 'Pandas'] | 0 |
get-the-size-of-a-dataframe | 🔥Best Solution in Pandas || With explanation ||🔥 | best-solution-in-pandas-with-explanation-qdpo | Explanation\ndef getDataframeSize(players: pd.DataFrame) -> List[int]:\nThis line defines a function called "getDataframeSize". It takes one argument players, w | deleted_user | NORMAL | 2024-03-16T23:07:45.112508+00:00 | 2024-03-16T23:07:45.112541+00:00 | 1,756 | false | # Explanation\n`def getDataframeSize(players: pd.DataFrame) -> List[int]:`\nThis line defines a function called "getDataframeSize". It takes one argument players, which is expected to be a pandas DataFrame. The function is annotated to indicate that players should be a DataFrame, and the return type of the function is ... | 6 | 0 | ['Python', 'Python3', 'Pandas'] | 1 |
get-the-size-of-a-dataframe | Pandas 1-Line | Elegant & Short | And more Pandas solutions | pandas-1-line-elegant-short-and-more-pan-oh92 | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Ankita2905 | NORMAL | 2023-11-08T17:07:33.838972+00:00 | 2023-11-08T17:07:33.839005+00:00 | 1,400 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 5 | 0 | ['Pandas'] | 1 |
get-the-size-of-a-dataframe | Easy Solution with simple explanation | easy-solution-with-simple-explanation-by-8zgc | Intuition\nWe need to determine the dimensions of the given DataFrame, which in pandas are easily accessible through built-in attributes.\n\n# Approach\nUse the | Swayam248 | NORMAL | 2024-09-27T15:42:17.225660+00:00 | 2024-09-27T15:42:17.225691+00:00 | 2,346 | false | # Intuition\nWe need to determine the dimensions of the given DataFrame, which in pandas are easily accessible through built-in attributes.\n\n# Approach\nUse the shape attribute of the DataFrame, which returns a tuple containing the number of rows and columns.\nConvert the tuple to a list to match the required output ... | 4 | 0 | ['Pandas'] | 6 |
get-the-size-of-a-dataframe | easy program | easy-program-by-pranov_raaj__30-lmko | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | pranov_raaj__30 | NORMAL | 2024-08-19T06:55:07.932119+00:00 | 2024-08-19T06:55:07.932155+00:00 | 2,095 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 4 | 0 | ['Pandas'] | 2 |
get-the-size-of-a-dataframe | easy to learn | easy-to-learn-by-yogesh_12-734h | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Yogesh_12__ | NORMAL | 2024-08-19T07:15:33.569474+00:00 | 2024-08-19T07:15:33.569501+00:00 | 513 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['Pandas'] | 0 |
get-the-size-of-a-dataframe | ✅EASY SOLUTION | easy-solution-by-swayam28-4edq | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | swayam28 | NORMAL | 2024-08-09T10:36:57.816018+00:00 | 2024-08-09T10:36:57.816056+00:00 | 876 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 3 | 0 | ['Pandas'] | 0 |
get-the-size-of-a-dataframe | ✔️✔️✔️easy 3 word solution - shape - PANDAS - beats 90%⚡⚡⚡ | easy-3-word-solution-shape-pandas-beats-m63pe | Code\n\nimport pandas as pd\ndef getDataframeSize(players: pd.DataFrame) -> List[int]:\n return list(players.shape)\n | anish_sule | NORMAL | 2024-04-10T05:23:03.540834+00:00 | 2024-04-10T05:23:03.540868+00:00 | 1,099 | false | # Code\n```\nimport pandas as pd\ndef getDataframeSize(players: pd.DataFrame) -> List[int]:\n return list(players.shape)\n``` | 3 | 0 | ['Pandas'] | 0 |
get-the-size-of-a-dataframe | Get the Size of a DataFrame - <PANDAS> | get-the-size-of-a-dataframe-pandas-by-pr-nxvb | Code | PreeOm | NORMAL | 2025-03-15T16:05:39.510839+00:00 | 2025-03-15T16:05:39.510839+00:00 | 630 | false |
# Code
```pythondata []
import pandas as pd
def getDataframeSize(players: pd.DataFrame) -> List:
return [players.shape[0], players.shape[1]]
``` | 2 | 0 | ['Pandas'] | 0 |
get-the-size-of-a-dataframe | Easy Explanation for Beginners ✨ | easy-explanation-for-beginners-by-amogha-eh9p | Intuitionshape is an attribute in pandas that returns a tuple (m,n).
where m is number of rows and n is number of columns.but the question asks us to return a l | AmoghaMayya | NORMAL | 2025-03-05T05:54:35.154897+00:00 | 2025-03-05T05:54:35.154897+00:00 | 590 | false | # Intuition
<!-- Describe your first thoughts on how to solve this problem. -->
**shape** is an attribute in pandas that returns a **tuple (m,n)**.
where **m** is number of rows and **n** is number of columns.
but the question asks us to return a list therefore we use **list()**.
we return a list in this way by making... | 2 | 0 | ['Pandas'] | 0 |
get-the-size-of-a-dataframe | 🚀1-row solution that beats 99999999,99%🚀 | 1-row-solution-that-beats-9999999999-by-0cwxi | ApproachTo solve this problem I read the pandas.DataFrame.shape documentation
df.shape return a tuple that consider the number of rows and number of columns (nr | cesar4ik | NORMAL | 2025-01-13T09:32:59.989202+00:00 | 2025-01-13T09:32:59.989202+00:00 | 1,323 | false | # Approach
To solve this problem I read the pandas.DataFrame.shape documentation
df.shape return a tuple that consider the number of rows and number of columns (nrows, ncols)
But we need to convert it to list to complete the problem condition
Goodbye, my firends!
<3
# Code
```pythondata []
import pandas as pd
def g... | 2 | 0 | ['Pandas'] | 0 |
get-the-size-of-a-dataframe | 🏆 Shape | 🥇 Beats 99.46% | ✨ Most Efficient | ✔ Explained | Python | shape-beats-9946-most-efficient-explaine-n7uc | Proof of 99.46%:\n\n\n# Code:\nPython\nimport pandas as pd\n\ndef getDataframeSize(players: pd.DataFrame) -> List[int]:\n return list(players.shape)\n \n\ | shlok_t | NORMAL | 2024-11-29T14:10:47.902550+00:00 | 2024-11-29T14:11:31.667977+00:00 | 317 | false | # Proof of 99.46%:\n\n\n# Code:\n```Python\nimport pandas as pd\n\ndef getDataframeSize(players: pd.DataFrame) -> List[int]:\n return list(players.shape)\n \n```\n# Approach:\n\n1. **Importing Pandas:... | 2 | 0 | ['Python', 'Python3', 'Pandas'] | 1 |
get-the-size-of-a-dataframe | Easy 1-line solution using .shape | easy-1-line-solution-using-shape-by-hiya-atfv | Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time | Hiya99 | NORMAL | 2024-09-29T05:01:31.143533+00:00 | 2024-09-29T05:01:31.143556+00:00 | 1,411 | false | # Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ --... | 2 | 0 | ['Pandas'] | 0 |
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